Problem: $s(t) = (t^4, t^4)$ What is the speed of $s(t)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(12t^2, 12t^2)$ (Choice B) B $(4t^3, 4t^3)$ (Choice C) C $t^4\sqrt{2}$ (Choice D) D $4t^3\sqrt{2}$
Solution: The speed of a parametric curve is the magnitude of its velocity. If $f(t) = (a(t), b(t))$, then speed is: $\| f'(t) \| = \sqrt{ a'(t)^2 + b'(t)^2 }$ Our position function here is $s(t)$. $\begin{aligned} s'(t) &= (4t^3, 4t^3) \\ \\ \text{speed} &= ||s'(t)|| \\ \\ &= \sqrt{16t^6 + 16t^6} \\ \\ &= 4t^3\sqrt{2} \end{aligned}$ Therefore, the speed of $s(t)$ is $4t^3\sqrt{2}$.